Weird Limits

Calculus Level 4

lim n ( 2 1000 n 1 ) n = ? \large \lim_{n\to\infty}\left(2\sqrt[n]{1000}-1\right)^{n} = \; ?


The answer is 1000000.

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2 solutions

Chew-Seong Cheong
Oct 17, 2017

Relevant wiki: L'Hôpital's Rule

L = lim n ( 2 1000 n 1 ) n = lim n ( 2 × 1 0 3 n 1 ) n = lim n exp ( ln ( 2 × 1 0 3 n 1 ) n ) where exp ( x ) = e x = exp ( lim n n ln ( 2 × 1 0 3 n 1 ) ) Let x = 1 n = exp ( lim x 0 ln ( 2 × 1 0 3 x 1 ) x ) A 0/0 case, L’H o ˆ pital’s rule applies. = exp ( lim x 0 2 ( 3 ln 10 ) 1 0 3 x 2 × 1 0 3 x 1 ) Differentiate up and down w.r.t. x = exp ( ln 1 0 6 ) = 1000000 \begin{aligned} L & = \lim_{n \to \infty} \left(2 \sqrt[n]{1000}-1\right)^n \\ & = \lim_{n \to \infty} \left(2\times 10^\frac 3n -1\right)^n \\ & = \lim_{n \to \infty} \exp \left(\ln \left(2\times 10^\frac 3n -1\right)^n \right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \exp \left( \lim_{n \to \infty} n \ln \left(2\times 10^\frac 3n -1\right) \right) & \small \color{#3D99F6} \text{Let }x = \frac 1n \\ & = \exp \left( \lim_{x \to 0} \frac {\ln \left(2\times 10^{3x} -1\right)}x \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \exp \left( \lim_{x \to 0} \frac {2(3\ln 10) 10^{3x}}{2\times 10^{3x} -1} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \exp \left( \ln 10^6 \right) \\ & = \boxed{1000000} \end{aligned}


Note:

f ( x ) = d d x ln ( 2 × 1 0 3 x 1 ) Let u = 2 × 1 0 3 x 1 = d d x ln u = 1 u × d u d x = 1 2 × 1 0 3 x 1 × d d x ( 2 × 10 3 x 1 ) Note that 10 = e ln 10 = 1 2 × 1 0 3 x 1 × d d x ( 2 e 3 x ln 10 1 ) = 2 ( 3 ln 10 ) e 3 x ln 10 2 × 1 0 3 x 1 = ( 6 ln 10 ) 1 0 3 x 2 × 1 0 3 x 1 \begin{aligned} f'(x) & = \frac {d}{dx} \ln ({\color{#3D99F6}2\times 10^{3x}-1}) & \small \color{#3D99F6} \text{Let } u = 2 \times 10^{3x} - 1 \\ & = \frac {d}{dx} \ln \color{#3D99F6} u \\ & = \frac 1u \times \frac {du}{dx} \\ & = \frac 1{2\times 10^{3x}-1} \times \frac d{dx} (2 \times {\color{#3D99F6}10}^{3x} - 1) & \small \color{#3D99F6} \text{Note that } 10 = e^{\ln 10} \\ & = \frac 1{2\times 10^{3x}-1} \times \frac d{dx} (2 {\color{#3D99F6}e}^{3x\color{#3D99F6}\ln 10} - 1) \\ & = \frac {2(3\ln 10)e^{3x\ln 10}}{2\times 10^{3x}-1} \\ & = \frac {(6\ln 10)10^{3x}}{2\times 10^{3x}-1} \end{aligned}

could you please explain and show the differentiation steps? I am somewhat confused as to how it was done.

Matthew Agona - 3 years, 7 months ago

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Note that d d x ( p x ) = p x ln p \dfrac{d}{dx}(p^x)=p^x\ln p

Dexter Woo Teng Koon - 3 years, 7 months ago

I have added an explanation as note.

Chew-Seong Cheong - 3 years, 7 months ago

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thank you!

Matthew Agona - 3 years, 7 months ago
Sibasish Mishra
Oct 21, 2017

this makes sense until right after the part where you show the term that tends to 1 as n approaches infinity. what did you do after that? where did the natural log go? how did you factor a 2 out?

Matthew Agona - 3 years, 7 months ago

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